In the trivial topology, where the only open sets are the entire space $X$ and the empty set $\emptyset$, let's address your questions:
$\textbf{Compactness:}$Every subset of $X$ is compact in the trivial topology. Compactness in a topology is not equivalent to boundedness in the metric space sense. In the trivial topology, compactness follows directly from the fact that any open cover for a subset must include the entire subset itself (if non-empty) or be empty.
$\textbf{Connectedness:}$A subset $A$ is connected in the trivial topology if the only way to express $A$ as the union of two disjoint open sets is to have one of them as $\emptyset$. For example, let's consider the set $[1, 3] \cup [4, 6]$ in the trivial topology on $\mathbb{R}$. In this topology, the only open sets are $\emptyset$ and $\mathbb{R}$. The set $[1, 3] \cup [4, 6]$ cannot be expressed as the union of two disjoint non-empty open sets. Therefore, it is connected in the trivial topology.
$\textbf{In summary:}$$\textbf{Compactness:}$ All subsets of $X$ are compact. $\\$$\textbf{Connectedness:}$ A subset is connected if it cannot be expressed as the union of two disjoint non-empty open sets in the trivial topology. For example, $[1, 3] \cup [4, 6]$ is connected in this topology.